Question

Calculate the uncertainty in position of an electron whose uncertainty in velocity is 0.1% of velocity of light

Answer 1

\(\Delta v = 0.1\% \times v\)

\(= \frac{1}{1000}\times 3 \times 10^8\)

⇒ \(3\times 10^5\)

we know that,

\(\Delta x\; \Delta p\ge \frac h{4\pi}\)

\(\Delta x\;m\,\Delta v = \frac h{4\pi}\)

\(\Delta x = \frac h{4\pi \,m\, \Delta v}\)

\(\Delta x = \frac {6.64\times 10^{-34}}{4\times 3.14\times 9.1 \times 10^{-31}\times 3 \times 10^5}\)

\(\Delta x = \frac{6.64\times 10^{-34}}{342.88 \times 10^{-26}}\)

⇒ \(0.01936\times 10^{-8} m\)